A particle is projected horizontally at 15ms\(^{-1}\) from a height of 20m. Calculate the horizontal...

Explanation

Let R represent the horizontal distance covered at time t.

Using h = ut + \(\frac{1}{2}gt^2\)

but U\(_v\) = 0

h = \(\frac{1}{2} gt^{2}\)

20 = \(\frac{1}{2} \times 10 \times t^2\)

20 = 5t\(^2\)

t\(^2\) = \(\frac{20}{5}\) = 4

t = 2s

Range(R) = U\(_x\) x t

= 15 x 2

= 30m

or

R = u x \(\sqrt{\frac{2h}{g}}\)

= 15 x \(\sqrt{\frac{2 \times 20}{10}}\)

= 15 x \(\sqrt{4}\)

= 15 x 2

= 30m